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Solve the ODE with initial condition:

  1. Feb 23, 2015 #1
    y''-10y'+25=0

    Solve the ODE with initial condition:

    y(0) = 0,

    y' (1) = 12e^5 .

    I keep getting y=12/5e^5x when c1=0 and c2=12/5 ... but Answer key says y=2xe^5x

    what am I doing wrong?
     
  2. jcsd
  3. Feb 23, 2015 #2
    We have to see what you're doing to know what you're doing wrong.
     
  4. Feb 23, 2015 #3
    I solved it and got general solution of y=c1e^5x+c2xe5x and the derivative is y'=c1(5e^5x)+c2(5xe^5x)
    for y(0)=0 i found that c1=0
    in y'(1)=12e^5 i found that c2=12/5

    which then gives final solution of y=(12/5)e^5x
     
  5. Feb 23, 2015 #4
    You need to use the product rule on the second term: [tex](x e^{5x})' = x' e^{5x} + x (e^{5x})' = e^{5x} + 5x e^{5x}.[/tex]

    Also, as a side note, it's advised that you use the homework section for any difficulties you're having in homework next time you have a question.
     
  6. Feb 23, 2015 #5

    Ohmygodd! You are right I did a silly mistake!

    This isnt homework, only studying for exam. But i thinkyes better to ask there.

    Thank you
     
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